Work done when kicking a soccer ball

[alingy1]

Homework Statement

We want to calculate the work done by kicking the soccer ball. The acceleration of the ball is 6m/s^2. The mass of the ball is 0,5 kg. The distance the ball travels is 20 meters. The distance that the foot kicks the ball is 10 cm.

Homework Equations

W=Fd
F=ma

The Attempt at a Solution

My question is whether we need to use the distance of 20m or 10 cm. The concept of work is new to me. I don't understand if the distance in the formula W=Fd refers to the distance that the force travels while being applied to the soccer ball. Or does it refer to the distance the ball travels (20m). Can you please explain why? I really want to fully understand the idea of work. W=Fd is just too vague for me, especially since I want to continue my studies in physics. Thanks.

 

[Dick]

Homework Statement

We want to calculate the work done by kicking the soccer ball. The acceleration of the ball is 6m/s^2. The mass of the ball is 0,5 kg. The distance the ball travels is 20 meters. The distance that the foot kicks the ball is 10 cm.

Homework Equations

W=Fd
F=ma

The Attempt at a Solution

My question is whether we need to use the distance of 20m or 10 cm. The concept of work is new to me. I don't understand if the distance in the formula W=Fd refers to the distance that the force travels while being applied to the soccer ball. Or does it refer to the distance the ball travels (20m). Can you please explain why? I really want to fully understand the idea of work. W=Fd is just too vague for me, especially since I want to continue my studies in physics. Thanks.

The d in W=Fd is the distance over which the force is applied. 10 cm.

 

[CWatters]

Once the ball leaves the foot the foot can't do any more work on the ball. So it's 10cm.

 

[HallsofIvy]

If an object with initial speed 0 has acceleration "a" m/s^2 for t seconds, then its final speed will be at and it will have gone a distance (1/2)at^2. Knowing that a= 6 m/s^2 and that it has gone 10 cm= .1 m, you can calculate the time required by solving (1/2)(6)t^2= 3t^2= 0.1. Use that time to find its final speed and then its kinetic energy. The work done is the same as the increase in energy.